I stumbled upon a transport equation of the form $$u_t(x,t)=u_x(x,t) + u_x(1,t).$$ Since I can write it in the form $u_t(x,t) = Lu(x,t)$ where L is some linear operator I thought that there must be some theory behind this type of equations. Unfortunately, so far my research was unsuccessful. Is there a way to solve the problem explicitly, e.g., by using method of characteristics in some way?

Thank you in advance!


$$u_t(x,t)-u_x(x,t) = u_x(1,t).$$ $u_x(1+t)$ is a function of $t$ only. Let $u_x(1+t)=f(t)$. Of course, this is an unknown function. $$u_t(x,t)-u_x(x,t) = f(t).$$ Solving for the general solution. Charpit-Lagrange equations : $$\frac{dt}{1}=\frac{dx}{-1}=\frac{du}{f(t)}$$ A first family of characteristic curves comes from $\frac{dt}{1}=\frac{dx}{-1}$ $$x+t=c_1$$ A second family of characteristic curves comes from $\frac{dt}{1}=\frac{du}{f(t)}$ $$u-\int f(t)dt=c_2$$ General solution of the PDE : $u-\int f(t)dt=\Phi(x+t)$

$\Phi(X)$ is an arbitrary function of one variable only. In the above equation $X=(x+t)$. $$u(x,t)= \Phi(x+t)+\int f(t)dt$$ $u_x(x,t)=\left(\frac{d\Phi(X)}{dX}\right)_{(X=x+t)}=\Phi'(x+t)$

$u_x(1,t)=\left(\frac{d\Phi(X)}{dX}\right)_{(X=1+t)}=\Phi'(1+t)$. $$f(t)=u_x(1,t)=\Phi'(1+t).$$ $\int f(t)dt=\int_{t_0}^t f(\zeta)d\zeta+C= \int_{t_0}^t \Phi'(1+\zeta)d\zeta +C =\Phi(1+t)+C$.

Finally the solution is : $$u(x,t)= \Phi(x+t)+\Phi(1+t)+C$$ $\Phi$ is an arbitrary function. $C$ is an arbitrary constant. They have to be determined to fit some boundary and initial conditions. One cannot go further since those conditions are not specified in the wording of the question.

CHECKING : $$u_x(x,t)=\Phi'(x+t)$$ $$u_t(x,t)=\Phi'(x,t)+\Phi'(1+t)$$ $$u_t(x,t)-u_x(x,t)=\Phi'(1+t)=u_x(1,t)$$ The PDE is satisfied. The above result is correct.

IN ADDITION after the comments :

Case of condition $u(x,0)=g(x)=\Phi(x)+\Phi(1)+C=\Phi(x)+C_2$


$u(x,t)= (g(x+t)-C_2)+(g(1+t)-C_2)+C$

Thus $\quad -2C_2+C+g(1)=0\quad;\quad C=2C_2-g(1)$



Example :

$$g(x)=x^m\qquad:\qquad u(x,t)=(x+t)^m+(1+t)^m-1$$

  • $\begingroup$ Thank you for your response. That was really helpful. In general my initial condition will just look like this: $u(x,0)=g(x)$ where g is some continuous function. For the sake of my example I choose $g(x) = x^m$ for some natural number $m$. Setting $C=-g(1)$ and $\Phi=g$ would yield a solution, since $$u(x,0)=\Phi(x+0) + \Phi(1) - g(1) = \Phi(x) = g(x)$$ and by your calculation. Is that right? But this choice is always possible independent of what my initial condition looks like as long as $g(1)$ is well defined and that is what puzzles me a bit. $\endgroup$ – Jfischer Nov 15 '18 at 8:13
  • $\begingroup$ I hoped to get the general concept from that toy example, i.e., how to solve an equation of type $$u_t(x,t) = w(x)u_x(x,t) + v(x)u_x(1,t)$$. I struggle with the step from $c_1 = x+t$ and $c_2 = u-\int f(t) dt$ to $u-\int f(t) dt = \Phi(x+t)$. Does that mean that $c_2$ is represented as a function along the characteristic? I would also be very thankful for a source for the case where a term of type $u_x(1,t)$ appears to read it up. $\endgroup$ – Jfischer Nov 18 '18 at 8:10

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